Prove the integral of 1/(x^3+8) between the x terminals 4 and 1 equals (pie root 3)/36

You can express,
As,
Then decompose this fraction as,

When fully decomposed gives,

The first summand calculates to,
Multiply by constant 1/8 and 2/3,.

The second summand need manipulation,

Apply a linear substitution,
Express as,

Yields, (first summand is natural logarithm in disuise, the second is the arctanget),
The first summand gives,
The second summand gives,
The next step is to subtract these summands,
Next is to mutiply this by -1/8 thus,
Now mutiply to 2/3,
And add your very first summand from the beginning,
thus,